MATH. SCAND. 93 (2003), 109–135 ON NON-HOLOMORPHIC FUNCTIONAL CALCULUS FOR COMMUTING OPERATORS SEBASTIAN SANDBERG Abstract We provide a general scheme to extend Taylor’s holomorphic functional calculus for several commuting operators to classes of non-holomorphic functions. These classes of functions will depend on the growth of the operator valued forms that define the resolvent cohomology class. The proofs are based on a generalization of the so-called resolvent identity to several commuting operators. 1. Introduction Let X, Y be two Banach spaces. We denote by L(X, Y ) the Banach space of all continuous linear operators from X to Y and we let L(X) = L(X, X).We denote by e the identity operator of L(X). For a subset A ⊂ L(X) we let A denote the bicommutant, that is the Banach algebra of all operators in L(X) which commute with every operator in {b ∈ L(X) : ab = ba for all a ∈ A} . Suppose that a ∈ L(X). The spectrum of a is defined as σ(a) = {z ∈ C : z − a is not invertible} , where z − a is the operator ze − a.Iff is a holomorphic function in a neigh- borhood of σ(a)then one can define the operator f(a)by the integral 1 (1.1) f (a) = f (z)(z − a)−1 dz, 2πi ∂D where D is an appropriate neighborhood of σ(a). This expression defines a continuous algebra homomorphism f → f(a): O(σ (a)) → (a), such that 1(a) = e and z(a) = a, called the Riesz functional calculus. We want to extend this algebra homomorphism to functions not necessarily holomorphic Received August 24, 2001. 110 sebastian sandberg in a neighborhood of the spectrum. Following Dynkin [6] we define f(a)by 1 (1.2) f (a) =− ∂f¯ (z) ∧ (z − a)−1 dz 2πi f ∈ S for all a, where ∂f ∂f¯ = dz¯ ∂z¯ and Sa is defined by S = f ∈ C1( ) f = ∂f¯ (z) ∧ (z − a)−1 dz < ∞ . a c C : a : ∞ It is evident that f(a) is a bounded linear operator on X which commutes with each operator that commutes with a, that is f(a) ∈ (a). By Stokes theorem the definition of f(a)only depends on the behavior of f near σ(a). Suppose that D is an open set such that σ(a) ⊂ D and that f ∈ O(D). Then 1 if φ ∈ Cc (D) is equal to 1 in a neighborhood of σ(a), we have that φf ∈ Sa and φf (a) defined by (1.2) equals f(a)defined by (1.1). We now prove that f → f(a) is an algebra homomorphism and that the spectral mapping theorem holds for functions in Sa. This is done in Dynkin [6], but we provide proof as we will generalize these theorems to the several oper- ator case. Theorem 1.1. The mapping f → f(a): Sa → (a) , where a ∈ L(X), is a continuous algebra homomorphism that continuously ex- tends the holomorphic functional calculus. Moreover, if f ∈ Sa then σ(f(a)) = f(σ(a)). Proof. The map f → f(a)is obviously linear and continuous. We have the so-called resolvent identity, (1.3)(w− z)(z − a)−1(w − a)−1 = (z − a)−1 − (w − a)−1 where z, w ∈ C. The multiplicative property then follows, f(a)g(a) 1 = ∂f¯ (z) ∧ (z − a)−1 dz ∧ ∂g(w)¯ ∧ (w − a)−1 dw (2πi)2 z w 1 = ∂f¯ (z) ∧ (z − a)−1 dz ∧ ∂g(w)¯ ∧ (w − z)−1 dw (2πi)2 z w on non-holomorphic functional calculus for commuting ... 111 1 + ∂f¯ (z) ∧ (z − w)−1 dz ∧ ∂g(w)¯ ∧ (w − a)−1 dw (2πi)2 z w 1 =− g(z)∂f¯ (z) ∧ (z − a)−1 dz πi 2 z 1 − f(w)∂g(w)¯ ∧ (w − a)−1 dw = fg(a), 2πi w by Fubini-Tonelli’s theorem. Suppose that D is an open neighborhood of σ(a)and that fn ∈ O(D) is a 1 sequence such that fn → 0 uniformly on compacts. Then if φ ∈ Cc (D) is a function equal to 1 in a neighborhood of σ(a)we have that fnφ a → 0. Thus the mapping f → f(a) continuously extends the holomorphic functional calculus. 1 If w/∈ f(σ(a))and φ ∈ Cc (C) is equal to 1 in an appropriate neighborhood of g(σ(a)), then φ ∈ S , w − f a and hence w − f(a)is invertible and thus w/∈ σ(f(a)). Therefore we have the inclusion σ(f(a)) ⊂ f(σ(a)). Suppose that w ∈ f(σ(a)) and assume that w = 0. Then 0 = f(ζ)for some ζ ∈ σ(a). Let f(z) g(z) = . z − ζ Then 1 f(a)=− (z − ζ)∂g(z)¯ ∧ (z − a)−1 dz 2πi z 1 = (ζ − a) ∂g(z)¯ ∧ (z − a)−1 dz 2πi z 1 − (z − a)∂g(z)¯ ∧ (z − a)−1 dz. 2πi z The last integral equals f(ζ), which is 0, and hence 0 ∈ σ(f(a))since other- wise ζ − a would be invertible. Therefore f(σ(a))⊂ σ(f(a)), and hence the theorem is proved. Furthermore, we have a rule of composition for this functional calculus. Theorem 1.2 (Rule of composition). If g ∈ Sa and f is a holomorphic function in a neighborhood of σ(a), then φ (f ◦ g) ∈ Sa and f(g(a)) = 1 φ (f ◦ g) (a),ifφ ∈ Cc (C) is equal to 1 in a neighborhood of σ(a). 112 sebastian sandberg 1 Proof. Suppose that ψ ∈ Cc (C) is equal to 1 in a neighborhood of σ(g(a)). 1 There is a function φ ∈ Cc (C) such that φ is equal to 1 in a neighborhood of σ(a)and φ h = ∈ S w − g a ¯ for each fixed w ∈ supp ∂ψ . The function φ(f ◦ g) is in Sa since ∂ (φ (f ◦ g)) ∂φ ∂f ∂g = f ◦ g + φ . ∂z¯ ∂z¯ ∂w ∂z¯ Since 1 ¯ −1 f(g(a))=− f(w)∂wψ(w) ∧ (w − g(a)) dw 2πi w 1 ¯ ¯ −1 = f(w)∂wψ(w) ∧ dw ∧ ∂zh(z) ∧ (z − a) dz (2πi)2 w z φ(z)dw 1 ¯ ¯ −1 = ∂z f(w)∂wψ(w) ∧ ∧ (z − a) dz (2πi)2 z w w − g(z) 1 ¯ −1 =− ∂z (φf ◦ g) ∧ (z − a) dz = φ(f ◦ g)(a) 2πi z the theorem is proved. For further results regarding this functional calculus, see Dynkin [6]. Now we turn to the notion of spectrum of a commuting tuple of operators. n Suppose that a = (a1,...,an) ∈ L(X) is a commuting tuple of operators, that is ai aj = aj ai for all i and j. Denote by ! =⊕n !p p=0 n n the exterior algebra of C over C.Ifs1, ···,sn is a basis of C then ! has the basis s = ,s= s ∧···∧s ,I={i , ···,i }, ∅ 1 I i1 ip 1 p where i1 < ··· <ip and 1 ≤ p ≤ n, and we denote ! = !(s) in this case. We let K•(a, X) be the Koszul complex induced by a, δp+1 δp ···→Kp+1(a, X) −−−→ Kp(a, X) −−→ Kp−1(a, X) →···, where p p Kp(a, X) = ! (s, X) = X ⊗C ! (s) on non-holomorphic functional calculus for commuting ... 113 and p δ (xs ) = πi (− )k−1a xs ∧···∧s ∧···∧s . p I 2 1 ik i1 ik ip k=1 If K•(a, X) is exact then a is called nonsingular, otherwise singular. The spec- trum is defined as σ(a) = z ∈ Cn : z − a is singular . One also defines the split spectrum as n sp(a) = z ∈ C : K•(z − a,X) is not split , where split means that for every integer p there are operators h and k such that e = δp+1h+kδp.IfX is a Hilbert space or n = 1 then sp(a) = σ(a). In general we have that σ(a) ⊂ sp(a), but not the reverse inclusion, see Müller [11]. We will consider operators parametrized by a variable z, such as z → z−a. In that case the boundary map δp depends on z and we will henceforth suppress the index p and write δp as δz−a for every p. We also let si = dzi . Now suppose that T ∈ L(X, Y ) has closed range and let k(T ) be the norm of the inverse of T considered as a map from X/ Ker T to T . The next lemma is Lemma 2.1.3 of [7], and it implies that if a0 is a nonsingular tuple then a is nonsingular if a0 − a is small enough. Lemma 1.3. Suppose that X, Y, Z are Banach spaces, α0 ∈ L(X, Y ), β0 ∈ L(Y, Z), β0 closed and Ker β0 =α0, that is α β X −−→0 Y −−→0 Z is exact. Let r be a number such that r>max {k(α0), k(β0)} .Ifα ∈ L(X, Y ), β ∈ L(Y, Z), α ⊂ Ker β and α − α0 , β − β0 < 1/6r then α = Ker β and k(α) ≤ 4r. Hence σ(a) is closed. Furthermore, the spectrum has the projection prop- erty, see Theorem 2.5.4 of [7]. n n+1 Theorem 1.4. If a ∈ L(X) and a = (a, an+1) ∈ L(X) are commuting n+1 n and π : C → C is defined by π(z,zn+1) = z then π(σ(a )) = σ(a). It follows that σ(a) ⊂ σ(a1) ×···×σ(an) and hence σ(a)is bounded. Thus σ(a)is a compact subset of Cn. Conversely, any compact set K in Cn can arise as the spectrum of a commuting tuple of 114 sebastian sandberg operators.